Question

(3r-2k)³+(3r+2k)³ SIMPLIFY ​

Answer 1

Answer:

126k^2r is the answer

Step-by-step explanation:

(3r-2k)3 + (3r+2k) 3

= {(3r-2k)+(3r+2k)} [(3r-2k)^2 -(3r-2k)(3r+2k) +(3r+2k) ^2]

= 6r. (9r^2 - 12kr + 4k^2 -9r^2 +6kr -6kr+4k^2 +9r2^ +12kr +4k^2)

= 6r. 21k^2

= 126k^2r

hope my answer helps

Answer 2

Answer:

Let us expand the given expression:

(3r – 2k)3 + (3r + 2k)3 = [(3r)3 -{3 ×(3r)2 ×(2k)} + {3 ×(3r)×(2k)2 } -(2k)3 ] + [(3r)3 +{3 × (3r)2 ×(2k)} + {3 ×(3r)×(2k)2} +(2k)3 ]

By using the formula,

(a + b)3 = a3 + 3a2b + 3ab2 + b3 and (a – b)3 = a3 – 3a2b + 3ab2 – b3

= [27r3 – {3 × 9r2 × 2k} + {3 × 3r × 4k2} – 8k3] + [27r3 + {3 × 9r2 × 2k} + {3 × 3r ×(4k2)} + 8k3]

= [27r3 – 54r2k + 36rk2 – 8k3] + [27r3 + 54r2k + 36rk2 + 8k3]

= 27r3 – 54r2k + 36rk2 – 8k3 + 27r3 + 54r2k + 36rk2 + 8k3

= 54r3 + 72rk2

∴ (3r – 2k)3 + (3r + 2k)3 = 54r3 + 72rk2